Question

With reference to the principal values, if sin^-1x + sin^-1y + sin^-1z = \(\frac {3π}{2}\), then x¹⁰⁰ + y¹⁰⁰ + z¹⁰⁰ =?

• A. 1
• B. 2
• C. 3
• D. 6

Answer 1

The Correct Option is C

sin^-1x + sin^-1y + sin^-1z = \(\frac {3π}{2}\)
We know that the principal values of sin^-1θ lie between -\(\frac {π}{2}\) and \(\frac {π}{2}\). Since the sum of the three angles is equal to \(\frac {3π}{2}\), it means that each angle must be equal to \(\frac {π}{2}\). Therefore, we have: 
sin^-1x = \(\frac {π}{2}\) 
sin^-1y  = \(\frac {π}{2}\)
sin^-1z  = \(\frac {π}{2}\)
Taking the sine of both sides of these equations: 
sin(sin^-1x) = sin \(\frac {π}{2}\)
sin(sin^-1y) = sin \(\frac {π}{2}\)
sin(sin^-1z) = sin \(\frac {π}{2}\)
Using the inverse sine function's property sin(sin^-1θ)  = θ, we simplify: 
x = 1 
y = 1 
z = 1 
Now, we can calculate the sum of their 100^th powers:
x¹⁰⁰ + y¹⁰⁰ + z¹⁰⁰ = 1¹⁰⁰ + 1¹⁰⁰ + 1¹⁰⁰ = 1 + 1 + 1 = 3 
Therefore, the value of x¹⁰⁰ + y¹⁰⁰ + z¹⁰⁰ is 3. 
Among the given options, (C) 3 is the correct answer.